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$2.76 \mathrm{~g}$ of silver carbonate on being strongly heated yield a residue weighing
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Verified Answer
The correct answer is:
$2.16 \mathrm{~g}$
$\begin{aligned} & 2 \mathrm{Ag}_2 \mathrm{CO}_3 \xrightarrow{\Delta} 4 \mathrm{Ag}+2 \mathrm{CO}_2+\mathrm{O}_2 \\ & 2 \times 276 \mathrm{gm} \quad 4 \times 108 \mathrm{gm}\end{aligned}$
■ $2 \times 276 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives $4 \times 108 \mathrm{gm}$
$\therefore 1 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives $=\frac{4 \times 108}{2 \times 276}$
$\therefore \quad 2.76 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives
$\frac{4 \times 108 \times 2.76}{2 \times 276}=2.16 \mathrm{gm}$
■ $2 \times 276 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives $4 \times 108 \mathrm{gm}$
$\therefore 1 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives $=\frac{4 \times 108}{2 \times 276}$
$\therefore \quad 2.76 \mathrm{gm}$ of $\mathrm{Ag}_2 \mathrm{CO}_3$ gives
$\frac{4 \times 108 \times 2.76}{2 \times 276}=2.16 \mathrm{gm}$
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